Solve 5-a=3a^2+1

$$5 - a = 3 a ^ { 2 } + 1$$

Solve for a

$$a = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$a=1$$

Steps Using Factoring By Grouping

Subtract 3a^{2} from both sides.

$$5-a-3a^{2}=1$$

Subtract 1 from both sides.

$$5-a-3a^{2}-1=0$$

Subtract 1 from 5 to get 4.

$$4-a-3a^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3a^{2}-a+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3a^{2}+aa+ba+4. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-3\times 4=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=3 b=-4$$

Rewrite -3a^{2}-a+4 as \left(-3a^{2}+3a\right)+\left(-4a+4\right).

$$\left(-3a^{2}+3a\right)+\left(-4a+4\right)$$

Factor out 3a in the first and 4 in the second group.

$$3a\left(-a+1\right)+4\left(-a+1\right)$$

Factor out common term -a+1 by using distributive property.

$$\left(-a+1\right)\left(3a+4\right)$$

To find equation solutions, solve -a+1=0 and 3a+4=0.

$$a=1 a=-\frac{4}{3}$$

Steps Using the Quadratic Formula

Subtract 3a^{2} from both sides.

$$5-a-3a^{2}=1$$

Subtract 1 from both sides.

$$5-a-3a^{2}-1=0$$

Subtract 1 from 5 to get 4.

$$4-a-3a^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3a^{2}-a+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -1 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$a=\frac{-\left(-1\right)±\sqrt{1+12\times 4}}{2\left(-3\right)}$$

Multiply 12 times 4.

$$a=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-3\right)}$$

Add 1 to 48.

$$a=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-3\right)}$$

Take the square root of 49.

$$a=\frac{-\left(-1\right)±7}{2\left(-3\right)}$$

The opposite of -1 is 1.

$$a=\frac{1±7}{2\left(-3\right)}$$

Multiply 2 times -3.

$$a=\frac{1±7}{-6}$$

Now solve the equation a=\frac{1±7}{-6} when ± is plus. Add 1 to 7.

$$a=\frac{8}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

$$a=-\frac{4}{3}$$

Now solve the equation a=\frac{1±7}{-6} when ± is minus. Subtract 7 from 1.

$$a=-\frac{6}{-6}$$

Divide -6 by -6.

$$a=1$$

The equation is now solved.

$$a=-\frac{4}{3} a=1$$

Steps for Completing the Square

Subtract 3a^{2} from both sides.

$$5-a-3a^{2}=1$$

Subtract 5 from both sides.

$$-a-3a^{2}=1-5$$

Subtract 5 from 1 to get -4.

$$-a-3a^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-3a^{2}-a=-4$$

Divide both sides by -3.

$$\frac{-3a^{2}-a}{-3}=-\frac{4}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$a^{2}+\left(-\frac{1}{-3}\right)a=-\frac{4}{-3}$$

Divide -1 by -3.

$$a^{2}+\frac{1}{3}a=-\frac{4}{-3}$$

Divide -4 by -3.

$$a^{2}+\frac{1}{3}a=\frac{4}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{1}{3}a+\left(\frac{1}{6}\right)^{2}=\frac{4}{3}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{1}{3}a+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Add \frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{1}{3}a+\frac{1}{36}=\frac{49}{36}$$

Factor a^{2}+\frac{1}{3}a+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{1}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$a+\frac{1}{6}=\frac{7}{6} a+\frac{1}{6}=-\frac{7}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$a=1 a=-\frac{4}{3}$$